Can you help me with an AP Physics question?

Question

This is the question: A rocket is fired vertically upward with an initial velocity of 80 m/s. It accelerates upward at 4 m/s^2 until it reaches an altitude of 1000m. At that point, its engines fail and the rocket goes into free flight, with acceleration -9.80 m/s^2. Now this is what I want to know: Will 1000m be the maximum altitude the rocket reaches since the engines stop at that moment, or will the rocket continue on? I know that at the maximum altitude, the velocity has to be 0 m/s, but according to my calculations using the data given, I get the final velocity of 120 m/s at 1000m. Please help!

Answer 1

I can tell you this much: 1000 m will not be the final altitude, as the rocket still has upward velocity when the engines fail. I can't tell you what its final altitude would be, as my maths are a little rusty.

Answer 2

No the Max Height will not be 1000m.
And yes the Velocity at 1000m will be 120m/s.
As far as i've calculated the Max Height reached by the Rocket will approxly be 1720m.
But gotto remember that practically the Rocket will not reach till that Height,since in Newtonian Mechanics we dont consider the various minute Variations which brings about the Changes in observed Result.

Answer 3

Initial velocity: 80m/s.
Then, movement uniformly accelerated at 4m/s2 until reaches 1000m
-> find the time it takes to run 1000m
e = a * t^2 / 2
or t = sqr(2 * e / a) = sqr (2 * 1000 / 4) = sqr(500) = 22.36 s
Now calculate the velocity at the instant the engine stops:
V = Vs + (a * t) = 80 + (4 * 22.36) = 169.44 m/s
From that speed, it is going to slow down by 9.81m/ss
Since V = a * t, we find t = V / a = 169.44 / 9.81 = 17.272 s
During these 17.272 sec, the rocket will travel
e = a * t^2 / 2 = 9.81 * 17.272^2 / 2 = 1463m
The total height attained is 1000m + 1463 = 2463m
(if my calculations are correct)

Answer 4

let u1 be initial velocity of object at ground level
Let v1 be velocity at height of 1000m above ground, on rise
Let a1 be the net acceleration on the rocket on rise = 4m/sec^2

from general equation: v1 ^ 2 = u1 ^ 2 + 2 * a1 * 1000
v1 ^2 = 80^2 + 2 * 4 * 1000 m/sec^2
v1 = 120 m/sec

At 1000m acceleration assumes a net downward acceleration of
9.8 m/Sec^2, applying the same eqn v^2 = u^2 + 2*a*s
v is zero at max height, acceleration is negative,
therfore 0 = 120^2 - 2 * 9.8 * s
and 120^2 = 19.6 * s
s = 120^2/19.6 m = 734.69 m
maximum height reached is 1734.69 m

Answer 5

no 1000m will not be the maximum altitude. u have to find the maximum height using 120 u find the time it takes to start its descend then u use the time to find the distance total distance and when u do that find velocity final 2 at the maximum height

Answer 6

NO. VELOCITY SHOULD BE ZERO. otherwise, the rocket would continue flying. But since its ACCELERATION IS given at -9.8 m/s^2, we can solve for the maximum height...

V2 =0 m/s
V1=120 ms/s
a= -9.8 m/s^2

v2^2=v1^2+2ad

0=(120)^2+2(-9.8)d

14400=19.6d

d=734.69

735+1000=1735 m

its maximum height must be 1735 m

Answer 7

erm dude when the engines failed it ll continue moving upwards, tink of a car traveling upslpoe at 300km/h den engine die, it ll still move up slightly before sliding down.. cos its accelerating downwards but it still has an upward initial velocity.. as for calculations i believe they hav done it for u le.. juz wanna get u to understand y is it so....=)

Answer 8

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Answer 9

I finished physics ages ago and don't feel like doing any questions. I'll give you tips though:
Treat vertical and horizontal separately.
Use suvat equations.

Hope you can work it out.